In triangle ABC, ∠A=60∘,∠B=40∘,∠A=60∘,∠B=40∘, and ∠C=80∘.∠C=80∘.
If P is the center of the circumcircle of the triangle ABC with radius unity, then the radius of the circumcircle of triangle $B P C$ is?

1

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\sqrt{3}

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2

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\frac{\sqrt{3}}{2}

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Solution

The correct option is B $1$
In $△ P B C$ , $∠ B O C = 120^{\circ}$ and $P B = P C = R = 1$
$\Rightarrow ∠ P B C = 30^{\circ}$
From Sine rule: $\frac{P C}{sin ∠ P B C} = 2 R_{1}$
where $R_{1}$ is the circumradius of $Δ P B C$
Therefore, $R_{1} = \frac{P C}{2 sin 30^{\circ}} = 1$
Ans: A

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In a triangle $A B C ∠ A = 60^{\circ}, ∠ B = 40^{\circ}$ and $∠ C = 80^{\circ}$ If P is the centre of the circumcircle of triangle $A B C$ with radius unity, then the radius of the circumcircle of triangle $B P C$ is

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In triangle ABC, ∠A=60∘,∠B=40∘,∠A=60∘,∠B=40∘, and ∠C=80∘.∠C=80∘. If P is the center of the circumcircle of the triangle ABC with radius unity, then the radius of the circumcircle of triangle BPC is?

The correct option is B 1In △PBC, ∠BOC=120∘ and PB=PC=R=1⇒∠PBC=30∘From Sine rule: PCsin∠PBC=2R1 where R1 is the circumradius of ΔPBCTherefore, R1=PC2sin30∘=1Ans: A

In triangle ABC, ∠A=60∘,∠B=40∘,∠A=60∘,∠B=40∘, and ∠C=80∘.∠C=80∘.
If P is the center of the circumcircle of the triangle ABC with radius unity, then the radius of the circumcircle of triangle $B P C$ is?

The correct option is B $1$
In $△ P B C$ , $∠ B O C = 120^{\circ}$ and $P B = P C = R = 1$
$\Rightarrow ∠ P B C = 30^{\circ}$
From Sine rule: $\frac{P C}{sin ∠ P B C} = 2 R_{1}$
where $R_{1}$ is the circumradius of $Δ P B C$
Therefore, $R_{1} = \frac{P C}{2 sin 30^{\circ}} = 1$
Ans: A